Question

${H_2} + C{I_2} \to 2HCl$
If 2.02 g of ${H_2}$ and 35.45 g of $C{I_2}$ are allowed to react, what is the limiting reactant and what mass of HCl will be formed?
A. ${H_2}$ , 72.92 g
B. $C{I_2}$ , 36.46 g
C. ${H_2}$ , 36.46 g
D. $C{I_2}$ , 18.23 g

Answer 1

1 mole of ${H_2}$ reacts with 0.5 mole of $C{I_2}$ to give 1 mole of HCl, we will find the mass of 1 mole of HCl. Chlorine is completely consumed in the reaction.

Complete step by step answer:
Given reaction is,
${H_2} + C{I_2} \to 2HCl$
The quantities of both the reactants are known i.e. 2.02 g of ${H_2}$ and 35.45 g of $C{I_2}$ are allowed to react. In order to find out which reactant is the limiting reagent; we have to compare them to each other and this comparison is done in moles.
Mass of ${H_2}$ = 2.02 g
Mass of $C{I_2}$ = 35.45 g
We know, Number of moles = Mass / Molecular mass
Moles of ${H_2} = \dfrac{{2.02}}{{2.02}}$ = 1 mole
Moles of $C{I_2} = \dfrac{{37.45}}{2} = \dfrac{1}{2}$ = 0.5mole
1 mole of ${H_2}$ reacts with 1 mole of $C{I_2}$ to give 2 moles of HCl. But $C{I_2}$ has only 0.5 mole.
So, only 1 mole of HCl is formed.
Mass of 1 mole of HCl = 1.01 + 35.45 = 36.46 g
So, the mass of HCl formed will be 36.46 g.
The reactant that is completely consumed in the reaction is the limiting reagent. The reaction stops when it is used, thus limiting the quantities of the products formed.
As chlorine is completely consumed, so $C{I_2}$ is the limiting reagent.

So, the correct answer is option (B).

Note: The ${H_2} + C{I_2} \to 2HCl$ reaction is the chemical combination reaction. In this reaction, 2 substances combine to form one substance.